Chemical Equilibrium (Lecture 7)

Introducing Reversible Reactions

Consider a chemical reaction:

$$a\mathrm{A}+b\mathrm{B}\stackrel{\mathrm{k}{}_1}{\to }c\mathrm{C}+d\mathrm{D}\text{\hspace{1em}(1)}$$

According to the rate law, the speed of this reaction is:

$$r_{\text{fwd}}=k_1[A{]}^a[B{]}^b\text{\hspace{1em}(2)}$$

So far, nothing new. Now, looking at the potential energy surface or even at the 2D reaction diagram, you might wonder: if molecules have enough kinetic energy to surpass the hill of potential energy moving from reactants to products, is there anything to stop them from moving in the reverse direction?

Sometimes there is (formation of a precipitate, evolution of a gas, formation of some incredibly strong compounds), but sometimes there isn’t. And when there’s nothing to stop the reaction from proceeding in the opposite direction, we write the reaction equation as something being reversible:

$$a\mathrm{A}+b\mathrm{B}\underset{\mathrm{k}{}_{-1}}{\overset{\mathrm{k}{}_1}{\rightleftharpoons }}c\mathrm{C}+d\mathrm{D}\text{\hspace{1em}(3)}$$

We might then write the rate of the reverse reaction as:

$$r_{\text{rev}}=k_{-1}[C{]}^c[D{]}^d\text{\hspace{1em}(4)}$$

In a reversible reaction, at any point in time, some reactants are converted into products, and some products are converted back to reactants.

we usually denote the rate constants of reverse reactions with negative indices just because it is a bit more informative. If you see $k_1$ and $k_{-1}$, it’s fair to assume that those are forward and reverse steps of some reactions. You wouldn’t be able to reach the same conclusion if we were to label those rate constants as $k_1$ and $k_2$ (just because they could be for a forward and reverse reaction, but they could also correspond to two totally different reactions)

A special moment in time

Are there any special states of such reversible system? Well, one particular case that could be of interest to you (doesn’t mean you could have suggested it, but it should be interesting in hindsight) is when the concentrations remain more or less constant. In other words, is there a point in time $\tau$ such that if we measure the concentration at that moment, and a moment later, at $t=\tau +1$, the following will be true?

$$[A{]}_{\tau }=[A{]}_{\tau +1}[B{]}_{\tau }=[B{]}_{\tau +1}[C{]}_{\tau }=[C{]}_{\tau +1}[D{]}_{\tau }=[D{]}_{\tau +1}\text{\hspace{1em}(5)}$$

(just to be clear, the above do not imply $[A{]}_{\tau }=[B{]}_{\tau }$, individual concentrations may be different, but each on its own is constant in time)

We can’t tell if such point will occur, but for now, let’s do some manipulations with conditions imposed above. If $[A{]}_{\tau }=[A{]}_{\tau +1}$, it means that $d[A]/dt=0$ (because $d[A]\approx \Delta [A]=[A{]}_{t+1}-[A{]}_t=0$). When would $\Delta [A]=0$? In case if the amount of A produced is equal to the amount of A consumed. A is consumed in the forward, reaction, for which we know:

$$r_{\text{fwd}}=k_1[A{]}^a[B{]}^b\text{\hspace{1em}(6)}$$

but we also know, by definition, that:

$$r_{\text{fwd}}=-\frac{1}{a}\frac{d[A]}{dt}=-\frac{1}{a}{\left(\frac{d[A]}{dt}\right)}_{\text{forward}}\text{\hspace{1em}(7)}$$

the labels in parentheses are added just to clarify that this is the $d[A]/dt$ which comes from the forward reaction. While we’re here, let’s note that this means:

$$\frac{1}{a}{\left(\frac{d[A]}{dt}\right)}_{\text{forward}}=-k_1[A{]}^a[B{]}^b\text{\hspace{1em}(8)}$$

Okay, now we also know A is produced in the reverse reaction. We know its rate:

$$r_{\text{rev}}=k_{-1}[C{]}^c[D{]}^d\text{\hspace{1em}(9)}$$

and we know that that rate, by definition is:

$$r_{\text{rev}}=\frac{1}{a}\frac{d[A]}{dt}=\frac{1}{a}{\left(\frac{d[A]}{dt}\right)}_{\text{reverse}}\text{\hspace{1em}(10)}$$

and again, let’s be explicit and say:

$$\frac{1}{a}{\left(\frac{d[A]}{dt}\right)}_{\text{reverse}}=k_{-1}[C{]}^c[D{]}^d\text{\hspace{1em}(11)}$$

we don’t have the minus sign here because A is a product! Now, the total change $d[A]$ should be the sum of the change $d[A]$ in the forward and reverse reactions:

$$\frac{1}{a}\frac{d[A]}{dt}=\frac{1}{a}{\left(\frac{d[A]}{dt}\right)}_{\text{forward}}+\frac{1}{a}{\left(\frac{d[A]}{dt}\right)}_{\text{reverse}}\text{\hspace{1em}(12)}$$

Let’s plug in the values from (8) and (11):

$$\frac{1}{a}\frac{d[A]}{dt}=-k_1[A{]}^a[B{]}^b+k_{-1}[C{]}^c[D{]}^d\text{\hspace{1em}(13)}$$

Remember that the condition (5) implied $\frac{d[A]}{dt}=0$? Then:

$$\begin{array}{r}\frac{1}{a}\frac{d[A]}{dt}=-k_1[A{]}^a[B{]}^b+k_{-1}[C{]}^c[D{]}^d=0⟹k_{-1}[C{]}^c[D{]}^d=k_1[A{]}^a[B{]}^b\end{array}\text{\hspace{1em}(14)}$$

Okay, so if the concentrations do not change over time, the following MUST be true:

$$k_{-1}[C{]}^c[D{]}^d=k_1[A{]}^a[B{]}^b\text{\hspace{1em}(15)}$$

which, according to (6) and (9) is equivalent to saying that $r_{\text{rev}}=r_{\text{fwd}}$, or the rate of the forward reaction is equal to the rate of the reverse reaction.

And we shall name this moment …

Your intuition might already tell you that the moment in time when (5) is satisfied is some kind of equilibrium. But if you don’t see why it has to be called equilibrium, we might as well simply define equilibrium to be the moment in time, when (5) is satisfied.

Then, (15) serves as the mathematical test for whether equilibrium is established.

To reiterate. We define the chemical equilibrium to be a state of the system in which the concentrations of each individual reactant and product do not change over time. In such state, the rates of the forward and reverse reactions are equal.

Deriving Equilibrium Constant

Let’s revisit equation (15). First, let’s recall that it doesn’t hold for any arbitrary values of $[A],[B],[C],[D]$, it only holds for some values. Let’s call the values of $[X]$ where $X=A,B,C,D$ for which (15) is true the equilibrium concentrations and denote them as $[X{]}_{eq}$. Then, we can rewrite (15) as:

$$k_{-1}[C{]}_{\text{eq}}^c[D{]}_{\text{eq}}^d=k_1[A{]}_{\text{eq}}^a[B{]}_{\text{eq}}^b\text{\hspace{1em}(16)}$$

we can rewrite it as:

$$\frac{[C{]}_{\text{eq}}^c[D{]}_{\text{eq}}^d}{[A{]}_{\text{eq}}^a[B{]}_{\text{eq}}^b}=\frac{k_1}{k_{-1}}\text{\hspace{1em}(17)}$$

We could have rewritten it as

$$\frac{[A{]}_{\text{eq}}^a[B{]}_{\text{eq}}^b}{[C{]}_{\text{eq}}^c[D{]}_{\text{eq}}^d}=\frac{k_{-1}}{k_1}\text{\hspace{1em}(18)}$$

as well, both are equivalent, and we have to pick one, so let’s say chemists agreed to chose the former way of writing things. Let’s forget about the left-hand side in (17). The RHS has:

$$\frac{k_1}{k_{-1}}$$

if you stare at this for a moment, you’ll notice that it doesn’t depend on concentrations (wow, thanks cap). Which means that once we pick a particular reaction, we’ll have some specific values of $k_1$ and $k_{-1}$ meaning that their ratio will be a constant value. It will be nice to give this ratio some name, so let us define the equilibrium constant $K$ as:

$$K\equiv \frac{k_1}{k_{-1}}\text{\hspace{1em}(19)}$$

the $\equiv$ sign means this equality is true by definition. According to (17), this means that $K$ is also equal to:

$$K=\frac{[C{]}_{\text{eq}}^c[D{]}_{\text{eq}}^d}{[A{]}_{\text{eq}}^a[B{]}_{\text{eq}}^b}\text{\hspace{1em}(20)}$$

Let’s imagine we measure concentrations of $A,B,C,D$ at some moment of time. If we calculate their ratio considering stoichiometric coefficients and call it $Q$:

$$Q=\frac{[C{]}^c[D{]}^d}{[A{]}^a[B{]}^b}\text{\hspace{1em}(21)}$$

then one way of looking at (20) is as an alternative mathematical test for equilibrium. If the value of $Q$ is equal to $K$, then we’re at equilibrium. One benefit of using $Q$ instead of (15) is that we can test for equilibrium without knowing the rate constants of forward and reverse reactions!